Calculus Examples

$y = \sin (x + \frac{π}{2})$

Step 1

Write $y = \sin (x + \frac{π}{2})$ as a function.

$f (x) = \sin (x + \frac{π}{2})$

Step 2

Find the first derivative of the function.

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = x + \frac{π}{2}$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $x + \frac{π}{2}$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [x + \frac{π}{2}]$

Step 2.1.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d x} [x + \frac{π}{2}]$

Step 2.1.3

Replace all occurrences of $u$ with $x + \frac{π}{2}$ .

$\cos (x + \frac{π}{2}) \frac{d}{d x} [x + \frac{π}{2}]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $x + \frac{π}{2}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{π}{2}]$ .

$\cos (x + \frac{π}{2}) (\frac{d}{d x} [x] + \frac{d}{d x} [\frac{π}{2}])$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\cos (x + \frac{π}{2}) (1 + \frac{d}{d x} [\frac{π}{2}])$

Step 2.2.3

Since $\frac{π}{2}$ is constant with respect to $x$ , the derivative of $\frac{π}{2}$ with respect to $x$ is $0$ .

$\cos (x + \frac{π}{2}) (1 + 0)$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $1$ and $0$ .

$\cos (x + \frac{π}{2}) \cdot 1$

Step 2.2.4.2

Multiply $\cos (x + \frac{π}{2})$ by $1$ .

$\cos (x + \frac{π}{2})$

Step 3

Find the second derivative of the function.

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = x + \frac{π}{2}$ .

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Step 3.1.1

To apply the Chain Rule, set $u$ as $x + \frac{π}{2}$ .

$f'' (x) = \frac{d}{d u} (\cos (u)) \frac{d}{d x} (x + \frac{π}{2})$

Step 3.1.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$f'' (x) = - \sin (u) \frac{d}{d x} (x + \frac{π}{2})$

Step 3.1.3

Replace all occurrences of $u$ with $x + \frac{π}{2}$ .

$f'' (x) = - \sin (x + \frac{π}{2}) \frac{d}{d x} (x + \frac{π}{2})$

Step 3.2

Differentiate.

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Step 3.2.1

By the Sum Rule, the derivative of $x + \frac{π}{2}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{π}{2}]$ .

$f'' (x) = - \sin (x + \frac{π}{2}) (\frac{d}{d x} (x) + \frac{d}{d x} (\frac{π}{2}))$

Step 3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - \sin (x + \frac{π}{2}) (1 + \frac{d}{d x} (\frac{π}{2}))$

Step 3.2.3

Since $\frac{π}{2}$ is constant with respect to $x$ , the derivative of $\frac{π}{2}$ with respect to $x$ is $0$ .

$f'' (x) = - \sin (x + \frac{π}{2}) (1 + 0)$

Step 3.2.4

Simplify the expression.

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Step 3.2.4.1

Add $1$ and $0$ .

$f'' (x) = - \sin (x + \frac{π}{2}) \cdot 1$

Step 3.2.4.2

Multiply $- 1$ by $1$ .

$f'' (x) = - \sin (x + \frac{π}{2})$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\cos (x + \frac{π}{2}) = 0$

Step 5

Take the inverse cosine of both sides of the equation to extract $x$ from inside the cosine.

$x + \frac{π}{2} = \arccos (0)$

Step 6

Simplify the right side.

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Step 6.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x + \frac{π}{2} = \frac{π}{2}$

Step 7

Move all terms not containing $x$ to the right side of the equation.

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Step 7.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$x = \frac{π}{2} - \frac{π}{2}$

Step 7.2

Combine the numerators over the common denominator.

$x = \frac{π - π}{2}$

Step 7.3

Subtract $π$ from $π$ .

$x = \frac{0}{2}$

Step 7.4

Divide $0$ by $2$ .

$x = 0$

Step 8

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x + \frac{π}{2} = 2 π - \frac{π}{2}$

Step 9

Solve for $x$ .

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Step 9.1

Simplify $2 π - \frac{π}{2}$ .

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Step 9.1.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x + \frac{π}{2} = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 9.1.2

Combine fractions.

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Step 9.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$x + \frac{π}{2} = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 9.1.2.2

Combine the numerators over the common denominator.

$x + \frac{π}{2} = \frac{2 π \cdot 2 - π}{2}$

Step 9.1.3

Simplify the numerator.

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Step 9.1.3.1

Multiply $2$ by $2$ .

$x + \frac{π}{2} = \frac{4 π - π}{2}$

Step 9.1.3.2

Subtract $π$ from $4 π$ .

$x + \frac{π}{2} = \frac{3 π}{2}$

Step 9.2

Move all terms not containing $x$ to the right side of the equation.

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Step 9.2.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$x = \frac{3 π}{2} - \frac{π}{2}$

Step 9.2.2

Combine the numerators over the common denominator.

$x = \frac{3 π - π}{2}$

Step 9.2.3

Subtract $π$ from $3 π$ .

$x = \frac{2 π}{2}$

Step 9.2.4

Cancel the common factor of $2$ .

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Step 9.2.4.1

Cancel the common factor.

$x = \frac{2 π}{2}$

Step 9.2.4.2

Divide $π$ by $1$ .

$x = π$

Step 10

The solution to the equation $x + \frac{π}{2} = \frac{π}{2}$ .

$x = 0, π$

Step 11

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin ((0) + \frac{π}{2})$

Step 12

Evaluate the second derivative.

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Step 12.1

Add $0$ and $\frac{π}{2}$ .

$- \sin (\frac{π}{2})$

Step 12.2

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$- 1 \cdot 1$

Step 12.3

Multiply $- 1$ by $1$ .

$- 1$

Step 13

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 14

Find the y-value when $x = 0$ .

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Step 14.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \sin ((0) + \frac{π}{2})$

Step 14.2

Simplify the result.

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Step 14.2.1

Add $0$ and $\frac{π}{2}$ .

$f (0) = \sin (\frac{π}{2})$

Step 14.2.2

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (0) = 1$

Step 14.2.3

The final answer is $1$ .

$y = 1$

Step 15

Evaluate the second derivative at $x = π$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin ((π) + \frac{π}{2})$

Step 16

Evaluate the second derivative.

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Step 16.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \sin (π \cdot \frac{2}{2} + \frac{π}{2})$

Step 16.2

Combine fractions.

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Step 16.2.1

Combine $π$ and $\frac{2}{2}$ .

$- \sin (\frac{π \cdot 2}{2} + \frac{π}{2})$

Step 16.2.2

Combine the numerators over the common denominator.

$- \sin (\frac{π \cdot 2 + π}{2})$

Step 16.3

Simplify the numerator.

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Step 16.3.1

Move $2$ to the left of $π$ .

$- \sin (\frac{2 \cdot π + π}{2})$

Step 16.3.2

Add $2 π$ and $π$ .

$- \sin (\frac{3 π}{2})$

Step 16.4

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$- - \sin (\frac{π}{2})$

Step 16.5

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$- (- 1 \cdot 1)$

Step 16.6

Multiply $- (- 1 \cdot 1)$ .

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Step 16.6.1

Multiply $- 1$ by $1$ .

$- - 1$

Step 16.6.2

Multiply $- 1$ by $- 1$ .

$1$

Step 17

$x = π$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = π$ is a local minimum

Step 18

Find the y-value when $x = π$ .

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Step 18.1

Replace the variable $x$ with $π$ in the expression.

$f (π) = \sin ((π) + \frac{π}{2})$

Step 18.2

Simplify the result.

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Step 18.2.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (π) = \sin (π \cdot \frac{2}{2} + \frac{π}{2})$

Step 18.2.2

Combine fractions.

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Step 18.2.2.1

Combine $π$ and $\frac{2}{2}$ .

$f (π) = \sin (\frac{π \cdot 2}{2} + \frac{π}{2})$

Step 18.2.2.2

Combine the numerators over the common denominator.

$f (π) = \sin (\frac{π \cdot 2 + π}{2})$

Step 18.2.3

Simplify the numerator.

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Step 18.2.3.1

Move $2$ to the left of $π$ .

$f (π) = \sin (\frac{2 \cdot π + π}{2})$

Step 18.2.3.2

Add $2 π$ and $π$ .

$f (π) = \sin (\frac{3 π}{2})$

Step 18.2.4

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$f (π) = - \sin (\frac{π}{2})$

Step 18.2.5

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (π) = - 1 \cdot 1$

Step 18.2.6

Multiply $- 1$ by $1$ .

$f (π) = - 1$

Step 18.2.7

The final answer is $- 1$ .

$y = - 1$

Step 19

These are the local extrema for $f (x) = \sin (x + \frac{π}{2})$ .

$(0, 1)$ is a local maxima

$(π, - 1)$ is a local minima

Step 20